The Circle C₁: x² + y² = 3, with centre at O, intersects the parabola x² = 2yat the point P in the first quadrant. Let the tangent to the circle C₁ at P touches other two circles C₂ and C₃ at R₂ and R₃, respectively.Suppose C₂ and C₃ have equal radii 2√3 and centers Q₂ and Q₃,respectively.If Q₂ and Q₃ lie on the y-axis then- [JEE Adv 2016 P1]

area of the triangle OR₂R₃ is 6{2}

area of the triangle PQ₂Q₃ is 4{2}

Test Index

Parabola

Section: Mathematics

Question : 15 of 31

Marks: +1, -0

The Circle

C_1: x^2 + y^2 = 3

x^2 = 2y

C_1

C_2

C_3

R_2

R_3

C_2

C_3

Q_2

Q_3

Q_2

Q_3

Solution:

On solving

x^2 + y^2 = 3

and x^2=2y we get point

(\sqrt{2},1)

Equation of tangent at P

\sqrt{2} \cdot x + y = 3

Let

Q_2

be (0,k) and radius is

2\sqrt{3}

\therefore \left| \frac{\sqrt{2}(0) + k - 3}{\sqrt{2} + 1} \right| = 2\sqrt{3}

\therefore k = 9, -3

Q_2(0,9)

and

Q_3(0,-3)

hence

Q_2, Q_3 = 12

R_2, R_3

is internal common tangent of circle

C_2

and

C_3

\therefore R_2, R_3 = \sqrt{ (Q_2, Q_3)^2 - (2\sqrt{3} + 2\sqrt{3})^2 }

= \sqrt{12^2 - 48} = \sqrt{96} = 4\sqrt{6}

Perpendicular distance of origin O from

R_2, R_3

is equal to radius of circle

C_1 = \sqrt{3}

Hence area of

\Delta OR_2R_3 = \frac{1}{2} \times (R_2, R_3) \sqrt{3} = \frac{1}{2} \cdot 4\sqrt{6} \cdot \sqrt{3} = 6\sqrt{2}

Perpendicular Distance of P from

Q_2 Q_3 = \sqrt{2}

∴Area of

\Delta PQ_2Q_3 = \frac{1}{2} \times 12 \times \sqrt{2} = 6\sqrt{2}

Go to Question:

Prev Question

Next Question